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September 13th, 2009, 05:54 PM   #1
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how can these to be factor?

3x^2 - 7x + 3

and

2x^3 + 128

i have no idea how the first one can be factor...
i dont think it can be factor...

the second one,
i went:
2[(x^3 + 64)]
could this be answer? seems like i could go more steps
haebin is offline  
 
September 14th, 2009, 06:02 AM   #2
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To learn how to factor quadratics (something they were supposed to have covered back in early algebra), try here.

For the second one, you need now to apply the difference-of-cubes factoring formula. (Memorize this, along with the sum-of-cubes formula and the difference-of-squares formula. You will need all of these!)
stapel is offline  
September 14th, 2009, 10:25 PM   #3
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We can possibly factor 3x - 7x + 3 using irrational numbers. Let the given expression equal to 0, so 3x - 7x + 3 = 0. Using quadratic formula for this equation, we arrive with roots of x = (7 ?(13)) / 6. Since 3x - 7x + 3 = 0 is equivalent to x - (7/3)x + 1 = 0, it's clear to see that (x - (7 + ?(13)) / 6)(x - (7 - ?(13)) / 6) = 0, then it's obvious that 3(x - (7 + ?(13)) / 6)(x - (7 - ?(13)) / 6) = 3x - 7x + 3 = 0. ? 3(x - (7 + ?(13)) / 6)(x - (7 - ?(13)) / 6).

To factor the expression 2x + 128 (or 2(x + 4)), use the suggested link given by stapel, we have 2(x + 4)(x - 4x + 16).
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